Given f(t) = t^2, g(t) = 2t^2 - 3t, u(t) = te_1 + t^(2)e_2 + 2te_3 and v(t) = e_1 - 2te_2 +...

Question:

Given

  • {eq}f(t) \, = \, t^{2}, \quad g(t) \, = \, 2t^{2} \, - \, 3t, \quad u(t) \, = \, te_1 \, + \, t^{2}e_2 \, + \, 2te_3 \quad {/eq} and {eq}\quad v(t) \, = \, e_1 \, - \, 2te_2 \, + \, 3t^{2}e_3 \, {/eq}

find the scalar of the components of the vectors

a. {eq}\quad \displaystyle{\frac{\mathrm{d} }{\mathrm{d} t} \, f(t) \, v(t)} {/eq}

b. {eq}\quad \displaystyle{\frac{\mathrm{d} }{\mathrm{d} t} \, f(t) \, u(t) \, \times \, v(t)} {/eq}

c. {eq}\quad \displaystyle{\int \, g(t) \, v(t) \, dt} {/eq}

Indefinite Integral:

There are several methods to solve the indefinite integral. The chain rule is one of them.

Let {eq}f\left( x \right) = {x^n} {/eq} be a function. The integral of the function can be defined as,

{eq}\int {f\left( x \right) = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c {/eq}

Here, {eq}c {/eq} is the integration constant.

Answer and Explanation:

Given

  • The functions are given by {eq}f\left( t \right) = {t^2},\;g\left( t \right) = 2{t^2} - 3t,\;u\left( t \right) = t{e_1} + {t^2}{e_2} + 2t{e_3},\;v\left( t \right) = {e_1} - 2t{e_2} + 3{t^2}{e_3} {/eq}.

(a)

Find the value of {eq}\dfrac{d}{{dt}}f\left( t \right)v\left( t \right) {/eq}.

First find the value of {eq}f\left( t \right)v\left( t \right) {/eq}.

{eq}\begin{align*} f\left( t \right)v\left( t \right)& = {t^2}\left( {{e_1} - 2t{e_2} + 3{t^2}{e_3}} \right)\\ & = {t^2}{e_1} - 2{t^3}{e_2} + 3{t^4}{e_3} \end{align*} {/eq}

Differentiate the function {eq}f\left( t \right)v\left( t \right) = {t^2}{e_1} - 2{t^3}{e_2} + 3{t^4}{e_3} {/eq} with respect to {eq}t {/eq}.

{eq}\begin{align*} \dfrac{d}{{dt}}\left[ {f\left( t \right)v\left( t \right)} \right] &= \dfrac{d}{{dt}}\left[ {{t^2}{e_1} - 2{t^3}{e_2} + 3{t^4}{e_3}} \right]\\ & = 2t{e_1} - 6{t^2}{e_2} + 12{t^3}{e_3} \end{align*} {/eq}

So, the value of {eq}\dfrac{d}{{dt}}f\left( t \right)v\left( t \right) is 2t{e_1} - 6{t^2}{e_2} + 12{t^3}{e_3} {/eq}.

(b)

Find the value of {eq}\dfrac{d}{{dt}}f\left( t \right)u\left( t \right) \times v\left( t \right) {/eq}

First find the value of {eq}f\left( t \right)u\left( t \right) {/eq}.

{eq}\begin{align*} f\left( t \right)u\left( t \right) &= {t^2}\left( {t{e_1} + {t^2}{e_2} + 2t{e_3}} \right)\\ & = {t^3}{e_1} + {t^4}{e_2} + 2{t^3}{e_3} \end{align*} {/eq}

Differentiate the function {eq}f\left( t \right)u\left( t \right) = {t^3}{e_1} + {t^4}{e_2} + 2{t^3}{e_3} {/eq} with respect to {eq}t {/eq}.

{eq}\begin{align*} \dfrac{d}{{dt}}\left[ {f\left( t \right)u\left( t \right)} \right] &= \dfrac{d}{{dt}}\left[ {{t^3}{e_1} + {t^4}{e_2} + 2{t^3}{e_3}} \right]\\ & = 3{t^2}{e_1} + 4{t^3}{e_2} + 6{t^2}{e_3} \end{align*} {/eq}

Find the cross product of {eq}\dfrac{d}{{dt}}f\left( t \right)u\left( t \right) {/eq} and {eq}v\left( t \right) {/eq}.

{eq}\begin{align*} \dfrac{d}{{dt}}f\left( t \right)u\left( t \right) \times v\left( t \right) &= \begin{vmatrix} e_1& e_2 &e_3 \\ 3t^2 & 4t^3 &6t^2 \\ 1 & -2t &3t^2 \end{vmatrix}\\ & = {e_1}\left( {12{t^5} + 12{t^3}} \right) - {e_2}\left( {9{t^4} - 6{t^2}} \right) + {e_3}\left( { - 6{t^3} - 4{t^3}} \right)\\ & = 12{t^3}\left( {1 + {t^2}} \right){e_1} - 3{t^2}\left( {3{t^2} - 2} \right){e_2} - 10{t^3}{e_3} \end{align*} {/eq}

So, the value of the {eq}\dfrac{d}{{dt}}f\left( t \right)u\left( t \right) \times v\left( t \right) {/eq} is {eq}12{t^3}\left( {1 + {t^2}} \right){e_1} - 3{t^2}\left( {3{t^2} - 2} \right){e_2} - 10{t^3}{e_3} {/eq}.

(c)

Find the value of {eq}\int {g\left( t \right)v\left( t \right)dt} {/eq} .

Find the value of {eq}g\left( t \right)v\left( t \right) {/eq}.

{eq}\begin{align*} g\left( t \right)v\left( t \right)&= \left( {2{t^2} - 3t} \right)\left( {{e_1} - 2t{e_2} + 3{t^2}{e_3}} \right)\\ & = 2{t^2}{e_1} - 4{t^3}{e_2} + 6{t^4}{e_3} - 3t{e_1} + 6{t^2}{e_2} - 9{t^3}{e_3}\\ & = \left( {2{t^2} - 3t} \right){e_1} + \left( {6{t^2} - 4{t^3}} \right){e_2} + \left( {6{t^4} - 9{t^3}} \right){e_3} \end{align*} {/eq}

Integrate the equation {eq}g\left( t \right)v\left( t \right) = \left( {2{t^2} - 3t} \right){e_1} + \left( {6{t^2} - 4{t^3}} \right){e_2} + \left( {6{t^4} - 9{t^3}} \right){e_3} {/eq}.

{eq}\begin{align*} \int {g\left( t \right)v\left( t \right)dt} & = \int {\left( {2{t^2} - 3t} \right){e_1} + \left( {6{t^2} - 4{t^3}} \right){e_2} + \left( {6{t^4} - 9{t^3}} \right){e_3}dt} \\ &= \left( {\dfrac{{2{t^3}}}{3} - \dfrac{{3{t^2}}}{2}} \right){e_1} + \left( {2{t^3} - {t^4}} \right){e_2} + \left( {\dfrac{{6{t^5}}}{5} - \dfrac{{9{t^4}}}{4}} \right){e_3} + c \end{align*} {/eq}

Here, c is integration constant.

Thus, the value of {eq}\int {g\left( t \right)v\left( t \right)dt} {/eq} is {eq}\left( {\dfrac{{2{t^3}}}{3} - \dfrac{{3{t^2}}}{2}} \right){e_1} + \left( {2{t^3} - {t^4}} \right){e_2} + \left( {\dfrac{{6{t^5}}}{5} - \dfrac{{9{t^4}}}{4}} \right){e_3} + c {/eq}.


Learn more about this topic:

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Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
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